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Finding energy eigenvalue of a harmonic oscillator using a Hamiltonian

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the energy eigenvalue.


    2. Relevant equations

    H = (p^2)/2m + 1/2m(w^2)(x^2) + λ(x^2)

    Hψ=Eψ

    3. The attempt at a solution

    So this is what I got so far:

    ((-h/2m)(∂^2/∂x^2)+(m(w^2)/2 - λ)(x^2))ψ=Eψ

    I'm not sure if I should solve this using a differential equation method, or is there an easier trick?

    Thank you!
     
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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    ... you should solve using the DE methods.

    Since the potential is in x only, the partials become exact[1]:
    $$\left ( \frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + \zeta x^2 - E \right )\psi(x) = 0 : 2\zeta = m\omega^2+2\lambda$$... and it is really useful for these things to learn to use ##\LaTeX## :)

    Note: You had:
    I don't know why you didn't square your ##\hbar## or why you have a ##-\lambda## in there but the setup has a ##+\lambda##.
    Also, I'm guessing by "##h##" you meant ##\hbar=h/2\pi## since the momentum operator is ##-i\hbar\frac{\partial}{\partial x}##.

    ----------------------

    [1] if you are expected to do this in 3D, then you'll have to deal with the other two directions.
     
    Last edited: Oct 11, 2012
  4. Oct 11, 2012 #3
    Ok, so now I tried to perform the differential equation, but then I get an equation in the form of:

    Ψ=Ae(ς-E)x+Be(-(ς-E)x)

    How can I find the energy eigenvalue from this equation?
     
  5. Oct 11, 2012 #4

    Simon Bridge

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    You put it into the Schrodinger equation and solve for E.

    But I am not sure that wavefunction is a solution... don't you associate Gaussians with harmonic oscillator potentials?
     
  6. Oct 11, 2012 #5
    We haven't done any sort of Gaussian...

    But I came up with another way to solve, can you see if this makes sense:

    I can write the Hamiltonian as:

    [itex] H =\frac{p^2}{2m} + x^2 (\frac{m\varpi^2}{2} + \frac{\lambda}{\sqrt{2}}) [/itex]

    or

    [itex] H = \frac{p^2}{2m} + \frac{mω'^2x^2}{2} [/itex]

    where

    [itex] ω'=\sqrt{ω^2+\frac{\sqrt{2}λ}{m}} [/itex]

    then I can use

    [itex] E = (\hbar)ω'(n+\frac{1}{2}) [/itex]
     
  7. Oct 11, 2012 #6

    Simon Bridge

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    Sure - you can exploit existing solutions. What what stopping you from writing ##m\omega^{\prime 2} = \zeta## before? (and using the same trick?)
     
  8. Oct 11, 2012 #7
    Good point! I can also do it that way, I'll try it out.

    This problem involves a 2D harmonic oscillator (which the Hamiltonian was for x). To find the degeneracy of the first excited state, can I state that since ωxy=ω, that n = nx + ny.

    Then I can write ny=n - x, and set up the following sum:
    [itex] \sum(n-n_{x}) = \frac{1}{2}n(n+1) [/itex]
    where the sum is from nx=0 to n.

    So, using the equation to solve for n=1 (first excited state), I get the degeneracy is equal to 1. Is this correct or am I missing something?
     
  9. Oct 11, 2012 #8

    Simon Bridge

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